Systems of equations and inequalities unit test answers pleaseee!!!

-4x+3y=-12 and -2x+3y=-18 is the second question.

The answer to this question is x=-6 and y=-3.

For the umpteenth time, the solution is wrong.

First equation.

- 4 x + 3 y = - 12

Multiply both sides by - 1

4 x - 3 y = 12

Add this equation to the second equation.

4 x - 3 y = 12
+
- 2 x + 3 y = - 18
_____________

2 x = - 6

x = - 6 / 2

x = - 3

Put this value in first equation.

- 4 x + 3 y = - 12

- 4 ∙ ( - 3 ) + 3 y = - 12

12 + 3 y = - 12

Subtract 12 to both sides.

3 y = - 24

y = - 24 / 3

y = - 8

The solution is:

x = - 3 , y = - 8

To solve the system of equations:

-4x + 3y = -12
-2x + 3y = -18

Step 1: Let's eliminate one variable by manipulating the two equations. We can do this by multiplying one equation by a number so that the coefficients of one variable are the same in both equations.

Step 2: Multiply the second equation by 2 to make the coefficients of x the same in both equations.

-4x + 3y = -12 (equation 1)
-2x + 3y = -18 (equation 2)

Multiply equation 2 by 2:
(-2x + 3y) * 2 = -18 * 2
-4x + 6y = -36 (equation 3)

Now we have two equations with the same coefficients for x.

Step 3: Subtract equation 1 from equation 3 to eliminate x.

(-4x + 6y) - (-4x + 3y) = -36 - (-12)
-4x + 6y + 4x - 3y = -36 + 12
3y = -24

Simplifying further, we have:
3y = -24

Step 4: Solve for y by dividing both sides of the equation by 3.

3y/3 = -24/3
y = -8

Step 5: Substitute the value of y back into one of the original equations to solve for x. Let's use equation 1.

-4x + 3(-8) = -12
-4x - 24 = -12

Step 6: Solve for x by isolating the x term.

-4x = -12 + 24
-4x = 12
x = 12/-4
x = -3

The solution to the system of equations -4x + 3y = -12 and -2x + 3y = -18 is x = -3 and y = -8.

I apologize, but I cannot provide specific answers to test questions. However, I can explain how to solve the given system of equations.

To solve the system of equations:
-4x + 3y = -12,
-2x + 3y = -18,

We can use the method of elimination or substitution.

Method 1: Elimination
1. Multiply both sides of the second equation by 2 to make the coefficients of x in both equations equal.
multiplying gives: -4x + 6y = -36.
2. Subtract the modified second equation from the first equation to eliminate x:
-4x + 3y - (-4x + 6y) = -12 - (-36).
Simplifying this equation gives: -4x + 3y + 4x - 6y = -12 + 36.
Simplifying further, we get -3y = 24.
3. Solve for y by dividing both sides of the equation by -3:
y = -8.
4. Substitute the value of y = -8 into any of the original equations to solve for x.
We will use the second equation: -2x + 3(-8) = -18.
Simplifying this equation gives: -2x - 24 = -18.
Add 24 to both sides: -2x = 6.
Divide through by -2: x = -3.

Therefore, the solution to the system of equations is x = -3 and y = -8.

Method 2: Substitution
1. Solve one equation for x or y in terms of the other variable.
From the first equation, we can solve for x:
-4x + 3y = -12,
-4x = -3y - 12,
x = (3y + 12)/(-4).
2. Substitute the expression for x into the second equation:
-2((3y + 12)/(-4)) + 3y = -18.
3. Simplify the equation by multiplying both sides by -4 to eliminate the fraction:
2(3y + 12) - 4(3y) = -18 * -4.
Simplifying further gives: 6y + 24 - 12y = 72.
4. Combine like terms on the left side of the equation:
-6y + 24 = 72.
5. Move the constant term to the right side by subtracting 24 from both sides:
-6y = 72 - 24,
-6y = 48.
6. Solve for y by dividing both sides by -6:
y = 48/(-6),
y = -8.
7. Substitute the value of y = -8 back into one of the original equations to solve for x.
Using the first equation: -4x + 3(-8) = -12,
-4x - 24 = -12.
Add 24 to both sides: -4x = 12.
Divide through by -4: x = -3.

Again, the solution to the system of equations is x = -3 and y = -8.